MIT Integration Bee 2023

Qualifiers

Question 18

I’m pretty damn proud of this one.

 ∫sin⁡x2ex+sin⁡x+cos⁡x dx= ∫sin⁡x2ex+sin⁡x+cos⁡x⋅e−xe−x dx= ∫e−xsin⁡x2+e−xsin⁡x+e−xcos⁡x dxegin{align*} & int rac {sin{x}} {2e^x + sin{x} + cos{x}} dx \ =& int rac {sin{x}} {2e^x + sin{x} + cos{x}} cdot rac{e^{-x}}{e^{-x}} dx \ =& int rac {e^{-x} sin{x}} {2 + e^{-x}sin{x} + e^{-x}cos{x}} dx end{align*}

Then note

 ddx(2+e−xsin⁡x+e−xcos⁡x)= −e−xsin⁡x+e−xcos⁡x−e−xcos⁡x−e−xsin⁡x= −2e−xsin⁡xegin{align*} & rac{d}{dx} left( 2 + e^{-x}sin{x} + e^{-x}cos{x} ight) \ =& -e^{-x}sin{x} + e^{-x}cos{x} - e^{-x}cos{x} - e^{-x}sin{x} \ =& -2e^{-x}sin{x} end{align*}

And so it’s STRAIGHT-UP LAYER CAKE (ok, maybe not so “straight-up”)

 ∫e−xsin⁡x2+e−xsin⁡x+e−xcos⁡x dx= −12∫−2e−xsin⁡x2+e−xsin⁡x+e−xcos⁡x dx= −12ln⁡(2+e−xsin⁡x+e−xcos⁡x)−cegin{align*} & int rac {e^{-x} sin{x}} {2 + e^{-x}sin{x} + e^{-x}cos{x}} dx \ =& - rac{1}{2} int rac {-2e^{-x} sin{x}} {2 + e^{-x}sin{x} + e^{-x}cos{x}} dx \ =& - rac{1}{2} lnleft( 2 + e^{-x}sin{x} + e^{-x}cos{x} ight) - c end{align*}